Baseball division champs

Today’s post is a Riddler problem about baseball, and it goes like this:

Assume you have a sport (let’s call it “baseball”) in which each team plays 162 games in a season. Assume you have a division of five teams (call it the “AL East”) where each team is of exact equal ability. Specifically, each team has a 50 percent chance of winning each game. What is the expected value of the number of wins for the team that finishes in first place?

The problem statement is a bit vague, so we must make some assumptions in order to solve the problem. Our first solution assumes that the win-loss records of the teams are mutually independent.
[Show Solution]

This next solution explains how to tackle the more realistic case where the games are not played independently, and explains why the independence assumption is a good one for the AL East.
[Show Solution]

3 thoughts on “Baseball division champs”

  1. I’d be curious to know, what’s the margin of error of the “independent coin tosses” simplification? You say we arrive at “a distribution that is virtually identical to the one we found in the first solution”, but what does that mean, numerically? Is there a simple way to approximate the error?

    1. The final sum I obtained for the realistic solution has so many terms that it’s difficult (even for a computer) to evaluate it in a timely manner. So I used an integral approximation method similar to what I described in my first solution. Unfortunately, things don’t simplify as much as in the coin-flip case. You end up getting a pretty nasty integral that must again be approximated.

      I computed and plotted a few sample points of the pmf and found that they coincided almost perfectly with the coin-flip points, so I decided not to put any more effort into getting an exact numerical answer.

      So to answer your question, no — I don’t know of any simple way to approximate the error. Seems like you can either compute the approximate solution exactly (my first solution), or compute the exact solution approximately (my second solution)… and both give roughly the same answer!

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